What's the derivative of $\int_1^x\sin(t)dt$?

What's the derivative of the integral $$\int_1^x\sin(t) dt$$

Any ideas? I'm getting a little confused.

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You can use the fundamental theorem of calculus, but if you have not yet covered that theorem, in short, you'll be taking the derivative - with respect to x - of the integral of $\sin(t)dt$ when the integral is evaluated from $1$ to $x$:

$$\frac{d}{dx}\left(\int_1^x \sin(t) \text{d}t\right) = \frac{d}{dx} [-\cos t]_1^x = \frac{d}{dx}\left(-\cos(x) - (-\cos(1))\right) = \sin(x).$$

and you'll no doubt be encountering the Fundamental Theorem of Calculus very, very soon:

For any integrable function $f$, and constant $a$: $$\frac{d}{dx} \left(\int_a^x f(t)dt \right)= f(x),$$

(provided $f$ is continuous at $x$).

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I think I've got it. d/dt(-cos(x)-(-cos(1)) = d/dt(-cos(x)+0) = sin(x) – Ian Nov 21 '12 at 19:35
Yes, you got it. – amWhy Nov 21 '12 at 19:36
Thanks for the help. – Ian Nov 21 '12 at 19:38
Your very welcome. – amWhy Nov 21 '12 at 19:41
Just be careful, $\cos(1) \neq 0$. – Emily Nov 21 '12 at 19:41

Using the fundamental theorem of calculus we know that the answer is $\sin(x)$

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$\frac{d}{dt}\int_1^x\sin(t)dt = \frac{d}{dt} [-\cos t]_1^x = \frac{d}{dt}[-\cos x+\cos(1)] = \sin x$

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If $f$ is any function at all that can be integrated, then the derivative of the integral of $f(t)dt$ from $1$ to $x$ is $f(x)$. This wonderful fact is the Fundamental Theorem of Calculus.

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I'm not so sure. For a trivial counter-example, if $\frac{d}{dx} \int_a^x f(t) dt = f(x)$ for every $x \in [a,b]$, then we can modify f on a set of measure 0 (say some countable set) and obtain a function for which this does not hold. The Fundamental theorem of Calculus holds in general only if f is continuous. – anonymous Nov 23 '12 at 1:39

You can use a nice theorem called the Fundamental Theorem of Calculus . Here, we're mainly worried about FTC part 1. Below is a summary of what FTC part 1 says.

Let $f$ be a continuous, function defined on $[a,b]$ and $$F(x) := \int_a^x{f(t)} \ dt \quad \quad \forall x \in [a,b]$$

Then, $F$ is continuous on the closed interval and differentiable on the open interval $(a,b)$ and $F'(x) = f(x) \ \forall x \in (a,b)$.

$$\frac{d}{dx}\int_1^x\sin(t) \ dt = \sin(x) \cdot \frac{d}{dx} (x) = \sin(x)$$
Note that you have to replace the $t's$ in the integrand with an $x$ and multiply by the derivative of the upper bound, assuming your lower bound is constant (which it is here.)
$$\frac{d}{dx}\int_1^{x^2}\sin(t) \ dt$$
This is the same as $$\sin(x) \cdot \frac{d}{dx}(x^2) = 2x \sin(x)$$